ArjanCodes’ Post

Most implementations of the State pattern in Python look very “clean”. Lots of small classes. A base interface. One class per state. But if you’ve ever worked with one in a real project, you know the downside: transitions are scattered, behaviour is hard to see in one place, and adding new states often means touching multiple files. In today’s video, I rebuild the State pattern in a very different way. Instead of relying on inheritance, I make the state machine explicit as data and use decorators to define transitions. The result is a small, reusable engine where the entire flow becomes visible at a glance. If you’re interested in writing Python that’s easier to reason about and extend, this is a pattern worth understanding. 👉 Watch here: https://lnkd.in/eg22yEHR. #python #softwaredesign #designpatterns #statemachine #cleancode

Day 42/100 – #100DaysOfCode 🚀 Solved LeetCode #2574 – Left and Right Sum Differences (Python). Today I practiced prefix sum logic to calculate the absolute difference between left and right sums for each index. Approach: 1) Calculate the total sum of the array. 2) Initialize leftSum = 0. 3) Traverse the array. 4) For each index, compute rightSum = total - leftSum - nums[i]. 5) Calculate the absolute difference and append it to the result. 6) Update leftSum by adding nums[i]. Time Complexity: O(n) Space Complexity: O(n) Understanding prefix sum helps solve problems efficiently 💪 #LeetCode #Python #DSA #Arrays #PrefixSum #ProblemSolving #100DaysOfCode

Day 3/365: Comparing Two Strings Character by Character 🧵🧠 Today I worked on a simple but fundamental logic problem: checking if two strings are the same, without directly using a built-in equality check. First, I compare the lengths of both strings. If lengths differ, they can’t be the same. If lengths match, I loop through each index and compare characters one by one. If any character is different, I break and print that the strings are not the same. If the loop finishes without finding a mismatch, the else block of the for loop runs and prints that the strings are the same. The interesting part is the for-else in Python. The else only runs when the loop completes normally (no break). This makes it a clean way to express: “if I didn’t find any mismatch in the entire loop, then the strings are equal.” Day 3 done ✅ 362 more to go. #100DaysOfCode #365DaysOfCode #Python #LogicBuilding #StringComparison #ForElse #CodingJourney #LearnInPublic #AspiringDeveloper

Day 5 of #30DaysOfPython ✅ Today I met two of Python's most powerful data structures. One of them already feels like home. The other? Slightly chaotic. Lists and dictionaries. Day 5. Lists made sense quickly — they're just ordered collections. I can store things, loop through them, sort them, slice them. Intuitive. Dictionaries? At first, the key-value pair concept felt abstract. The bug that got me today? I threw both strings and integers into the same list and tried to sort it. Python did not appreciate that. TypeError showed up like an old enemy. Day 5 done. 25 more to go! 👇 Lists vs dictionaries — when do you reach for one over the other? #Python #30DaysOfPython #DataStructures #StudentLife #AIML

Python Clarity Series – Episode 25 Topic: Mutable vs Immutable Function Behavior 📌 Why did my list change after function call? def modify(lst): lst.append(100) a = [1, 2] modify(a) print(a) Output: [1, 2, 100] 👉 Lists are mutable → changes reflect outside Now: def modify(x): x = x + 10 a = 5 modify(a) print(a) Output: 5 👉 Integers are immutable → no change outside 💡 Rule: Mutable → changes persist Immutable → changes don’t This confusion causes logic errors. #PythonBasics #FunctionConcepts #StudentClarity #python #clarity

Python Series — Day 3 🧠 Let’s level it up a bit 👇 What will be the output of this code? def modify_list(lst): lst.append(4) a = [1, 2, 3] modify_list(a) print(a) Options: A. [1, 2, 3] B. [1, 2, 3, 4] C. Error D. None Think carefully 👀 (Hint: It’s not about functions… it’s about how Python handles data) Drop your answer 👇 Answer tomorrow 🚀 #Python #CodingChallenge #LearningInPublic #DataEngineering #Tech

Ever tried to sort a list and ended up with None? The logic looks correct: nums = [3, 1, 2] sorted_nums = nums.sort() print(sorted_nums) 👉 Output: None Why did it fail? In Python, there is a big difference between a Method that modifies an object and a Function that returns a new one. 1️⃣ .sort() is a Method: It modifies the original list "in-place." It doesn't need to return anything because the work is done directly on the original variable. 2️⃣ sorted() is a Function: It creates a brand-new list and leaves the original one exactly as it was. Use .sort() when you want to save memory and don't need the original order anymore. Use sorted() when you need to keep your original data safe and want a new sorted version. #Python #30DaysOfCode #BCA #LearningInPublic #Day22 #JECRC

Python Clarity Series – Episode 23 Topic: Floating Point Precision Issue 🤯 Why does this happen? print(0.1 + 0.2) Output: 0.30000000000000004 ❗ 👉 This is NOT a Python bug. It’s due to how floating-point numbers are stored in binary. 💡 Fix (when needed): round(0.1 + 0.2, 1) Output: 0.3 💡 Concept: Computers approximate decimal values internally. Important in: ✔ Financial calculations ✔ Data Science Don’t ignore this. #PythonConcepts #FloatingPoint #RealWorldCoding #python #clarity

Day 7 of my Python learning journey: Today I focused on classic array and string patterns, and tried to keep solutions clean and efficient. What I solved: Two Sum using brute-force and hash map Valid Palindrome with two pointers Move Zeros with an in-place two-pointer approach Container With Most Water in O(n) Big takeaway: Correctness first, clarity second, optimization third. Small design choices, like in-place updates vs extra arrays, really affect code quality. GitHub link: https://lnkd.in/gGPw8_js #Python #ProblemSolving #Algorithms #TwoPointers #LearningInPublic #CleanCode

Day 55/100 – #100DaysOfCode 🚀 Solved LeetCode #205 – Isomorphic Strings (Python). Today I practiced hashmap (dictionary) usage to check whether two strings follow the same pattern. Approach: 1) Create two hashmaps to store character mappings in both directions. 2) Traverse both strings together using zip(). 3) Check if the current mapping is consistent in both maps. 4) If any mismatch is found, return False. 5) Otherwise, update the mappings and continue. 6) If all mappings are valid, return True. Time Complexity: O(n) Space Complexity: O(n) Understanding how bidirectional mapping ensures consistency 💪 #LeetCode #Python #DSA #HashMap #Strings #ProblemSolving #100DaysOfCode